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Physics Chp 13 SOUND. Compression vs Rarefaction Pitch – how we perceive the frequency Speed depends on medium (air, water, ice) 3D – goes in all directions.

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Presentation on theme: "Physics Chp 13 SOUND. Compression vs Rarefaction Pitch – how we perceive the frequency Speed depends on medium (air, water, ice) 3D – goes in all directions."— Presentation transcript:

1 Physics Chp 13 SOUND

2 Compression vs Rarefaction Pitch – how we perceive the frequency Speed depends on medium (air, water, ice) 3D – goes in all directions

3 Doppler Effect As you or the source moves the sound changes. Towards – higher Away - lower

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5 Sound Intensity I = P/A I = P/ 4πr 2 Threshold of hearing I o = 1x10 -12 W/m2

6 Audible Sounds Depends on intensity as well as frequency Resonance is when something vibrates at a natural frequency

7 Our Ear converts the sound wave into electrical waves so our brain can sense the sound.

8 Standing waves Simplest is the fundamental. On a string it has two nodes and one antinode. λ 1 =2L

9 For a string the first standing wave is half a wavelength and then goes up by whole intergers

10 Then it increases by additional antinodes λ 2 =L λ 3 = 2/3L λ 4 =1/2L

11 Since v=fλ and the speed remains const f n = n v/2L n= 1,2,3….. Harmonic series f 1 is the fundamental

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13 BUT a closed on one end is different Then the node has to hit the closed end to reflect back with no loss of the wave f n = n(v/4L) n= 1,3,5……

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15 If you use a mid C (261.6 hz) over a closed tube and get a resonant sound at 0.983 m and 1.64 m, which harmonics are present and what is the likely length for the first harmonic? Assume v = 343 m/s

16 f n = n(v/4L) n= 1,3,5…… 261.6 hz = n (343 m/s / 4(0.983m) ) n = 3 261.6 hz = n (343 m/s / 4(1.64m) ) n = 5 So if n is 1 then 261.6 hz = 1(343m/s / 4L) L = 0.328 m

17 When two frequencies are slightly off they interfere with each other and create a pattern of changing intensities. The difference in frequencies gives the beat frequency. f1 = 256hz and f2= 259 hz beat freq = 3hz


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